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Thread: Limit of a function that has a common denominator

  1. #1
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    Limit of a function that has a common denominator

    lim as x goes to 1 of

    (x^3 - 1)/(x^4 - 1)

    Evaluated as shown above, 0/0 is obtained, which leads me to believe there is a common factor between the divisor and the dividend that can be cancelled so the limit can be evaluated. The problem is, I have no idea how to deal with difference of cubes and difference of 4th degrees. Any help would be much appreciated.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by evankiefl View Post
    lim as x goes to 1 of

    (x^3 - 1)/(x^4 - 1)

    Evaluated as shown above, 0/0 is obtained, which leads me to believe there is a common factor between the divisor and the dividend that can be cancelled so the limit can be evaluated. The problem is, I have no idea how to deal with difference of cubes and difference of 4th degrees. Any help would be much appreciated.
    You can factor out x-1 and in both the numerator and the denominator, then cancel it:

    \frac{x^3-1}{x^4-1}=\frac{(x-1)\cdot(x^2+x+1)}{(x-1)\cdot (x^3+x^2+x+1)}=\ldots
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  3. #3
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    Quote Originally Posted by evankiefl View Post
    lim as x goes to 1 of

    (x^3 - 1)/(x^4 - 1)

    Evaluated as shown above, 0/0 is obtained, which leads me to believe there is a common factor between the divisor and the dividend that can be cancelled so the limit can be evaluated. The problem is, I have no idea how to deal with difference of cubes and difference of 4th degrees. Any help would be much appreciated.
    You could also use L'Hospital's Rule...

    \displaystyle \lim_{x \to 1}\frac{x^3 - 1}{x^4 - 1} = \lim_{x \to 1}\frac{\frac{d}{dx}(x^3 - 1)}{\frac{d}{dx}(x^4 - 1)}

    \displaystyle = \lim_{x \to 1}\frac{3x^2}{ 4x^3}

    \displaystyle = \frac{3}{4}.
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  4. #4
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    \displaystyle\frac{{{x}^{3}}-1}{{{x}^{4}}-1}=\frac{(x-1)\left( {{x}^{2}}+x+1 \right)}{(x+1)(x-1)\left( {{x}^{2}}+1 \right)}.
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  5. #5
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    Thanks everyone, and interesting L'Hopital's Rule technique. What exactly is the rule?
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  6. #6
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    With L'Hopital's Rule, whenever you have an indeterminant form (see below), you can take the derivative of the numerator, denominator, and then run the limit.

    Indeterminant forms:

    \displaystyle\frac{0}{0}, \ \ \frac{\infty}{\infty}, \ \ 0*\infty, \ \ 1^{\infty}, \ \ 0^0, \ \ \infty^0, \ \ \infty - \infty
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